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Solve the following Assignment problem (Hungarian Method) Max Z= 60 X1 + 5 X12 + 8 X13 + X14 + 2 X21 + [2 22 + 6 X23 + 5 X24 X31+ 8 X32 + 3 X33 + 9 X34 + 2 X41 + 4 ...

Question

Solve the following Assignment problem (Hungarian Method) Max Z= 60 X1 + 5 X12 + 8 X13 + X14 + 2 X21 + [2 22 + 6 X23 + 5 X24 X31+ 8 X32 + 3 X33 + 9 X34 + 2 X41 + 4 X42 + 6 X43 + 10 X44 s.t. X11 + X12 + X13 + X14 = X21 + X22 + X23 + X24 = X31 ~ X32 + X33 + X34 = X41 + X42 + X43 + X44 = X1l + X21 + X31+ X41 = X12 + X22 ~ X32 + X42 = X13 + X23 + X33 + X43 = X14 + X24 + X34+ X44 = xii = {0, 1}

Solve the following Assignment problem (Hungarian Method) Max Z= 60 X1 + 5 X12 + 8 X13 + X14 + 2 X21 + [2 22 + 6 X23 + 5 X24 X31+ 8 X32 + 3 X33 + 9 X34 + 2 X41 + 4 X42 + 6 X43 + 10 X44 s.t. X11 + X12 + X13 + X14 = X21 + X22 + X23 + X24 = X31 ~ X32 + X33 + X34 = X41 + X42 + X43 + X44 = X1l + X21 + X31+ X41 = X12 + X22 ~ X32 + X42 = X13 + X23 + X33 + X43 = X14 + X24 + X34+ X44 = xii = {0, 1}



Answers

Use a graphing calculator, Excel, or other technology to solve the following linear programming problems.
Maximize $\quad z=2.0 x_{1}+1.7 x_{2}+2.1 x_{3}+2.4 x_{4}+2.2 x_{5}$
subject to: $\quad 12 x_{1}+10 x_{2}+11 x_{3}+12 x_{4}+13 x_{5} \leq 4250$
$\quad 8 x_{1}+8 x_{2}+7 x_{3}+18 x_{4}+5 x_{5} \leq 4130$
$\quad 9 x_{1}+10 x_{2}+12 x_{3}+11 x_{4}+8 x_{5} \leq 3500$
$\quad 5 x_{1}+3 x_{2}+4 x_{3}+5 x_{4}+4 x_{5} \leq 1600$
with $\quad x_{1} \geq 0, \quad x_{2} \geq 0, \quad x_{3} \geq 0, \quad x_{4} \geq 0, \quad x_{5} \geq 0$

For this problem. We have been asked to find the dual problem for the given maximization Question, uh, that I've copied here for you to look at. We're supposed to maximize Z equals four X up one plus three, except two plus two, except three. And we want to know what the dual problem is for this. Now, duality says that every standard maximization problem has a corresponding associate ID standard minimization problem that have the same answer to it now we're not going to solve it. We just want to see what the dual pro problem is. So in order to set that up first, we're going to make a matrix with all of these coefficients. Okay, so let's start. We do this in green. Let's start with our conditions on. We're gonna put just the coefficients into an augmented matrix. Okay, so our first line one except 11 except to one except three less than or equal to five. So those were my coefficients and my constant terms. Next, I have 1104 There is no, except three. So we do have to put a zero there, and our last one is 213 and 15. Hey, now we're going to complete this matrix. The bottom row is going to be our maximization equation itself. I'm not going to be putting in negatives like we sometimes do with simplex method. Just the coefficients for three and to equals zero. Okay, now, next step, we're going to transpose this matrix. That means Rose become columns. Columns become rose. So take a look at this. First row here. 1115 We're now going to write that as a column instead of a row. 1115 Next second row 1104 We're going to write that as the second column. Okay, third row, 213 15. Third row becomes our third column, and our fourth row becomes our fourth column right now. From this, we can get our dual problem of a minimization function. So let's write this down here. We want to minimize. Okay? Are minimization function. We're going to get by looking at the coefficients on our bottom row here, so it's gonna be five. And because we have switched from maximization, the minimization I don't want to use X is I'm gonna use wise is my variables so five y sub one plus four y sub two plus 15 y sub three. And the convention your textbook is using is using a w here instead of a Z again showing that we are in a different, uh, problem set right now. Okay, Now, for our constraints were going to be subject thio the following constraints and we're gonna find thes by reading across our new columns. Kent, we're gonna use wise as our variables. So I have why sub one plus wise up to Plus to wise up three. Now, if if you look when we did our maximization we use less than or equal to for minimization we're going to switch that and use greater than or equal to four. Okay, second row y sub one plus wise up to plus wise up three greater than or equal to three and our last one. Why someone no y sub choose. Plus why Sub three is greater than or equal to two. And as always, we're going to say that thes variables do have to be greater than Sirrah. So this is our do a pro problem for our given maximization question,

In order to maximize the objective function, C equals five x plus three y. We have to graft the constraints to which it is subject to X and Y being greater than or equal to zero puts us in the first quadrant and the other constraints are graft and their corresponding colors. From these constraints, we confined the feasible points at the following circle locations. These are the points corresponding to the circled locations. In order to find which one of these points maximizes Z equals five x plus three y, we will plug in each one of these points in for X and y Doing so gives us the following values from these values we confined that the point that maximizes he is to six, giving us a value of 28 for Z.

For this problem. We have been given an equation Z and asked to maximize it. Now Z has five variables X up one through except five and we want to maximize given four constraining equations. We also know that we want all of our variables except one through except five to be positive, greater than or equal to zero. Now you can use any technology you would like in order to solve. This graphing calculator is a great tool you can use Excel. Um, there are several free online solutions as well. Like the one I found this website. As you can see, I've already typed in our equation maximize our equation z subject to the four constraining equations. And if I hit solve, I can look at our optimal solution and I'm just going to scroll over to see what we have. So we have a value of C equal to 795.68 and I'm gonna rounded to two decimal digits. So 795.68 And here the values for are my variables that give me that maximum value for Z except one will be zero and accept two will be zero as well. Except three is 46.97 except 4 176.72 and accept five is 124.5 So just gonna write those values down Z is 795.68 except one and ex of two were both zero except 3. 46.97 except 4 176.72 and accept five is 124.5 So these values for my five variables give me the maximum value for my function, see?


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